[tocomment]

Edit: I'm going to try to rephrase this whole question since it got so unpopular. (I also said moon when I meant sun)

Lets say we fly a rocket up to earth sun L4 at a really slow speed. When we reach L4 we fire retro rockets to slow down as to not overshoot.

It seems to me that we never have to reach a high speed to stay in space at that point. The two bodies would be holding us there?

Is that correct? Is there a certain speed to fly out to L4 which uses less fuel than speeding up to 8km/s like you would need to stay in space orbiting earth?

[simonh]

The most efficient way to get from low Earth orbit to a point further away from Earth is with a single burn of your rocket engine. This burn close to Earth launches your rocket on a ballistic curve to the target point, with the least energy on arrival at the target (slowest speed) necessary to get there. Necessarily, this means you start off going fast after the initial burn and you slow down due to Earth's gravity as your altitude increases.

Trying to do it more slowly will just waste fuel. It would mean burning your engine less when close to Earth, not enough to reach your destination, then burning it again at higher altitude to stop Earth's gravity pulling you back down again. To understand why this is less efficient, check out the Oberth Effect (1). Trying to get to a distant point like the Earth-Sun L4 point doesn't fundamentally change any of this.

(1) http://en.wikipedia.org/wiki/Oberth_effect

[MrMeker]

If you did that you would have a huge relative velocity to the Lagrange point. You'd shoot right past it unless you could kill all your relative velocity, which would mean you would have enough delta-v to achieve orbit anyway.

[tocomment]

I updated my question to explain it better I think.

[InclinedPlane]

It takes more speed to get high enough to reach an Earth-Moon lagrange point (L1 would be the closest, at about 90% of the Earth-Moon distance) than to simply get into Earth orbit.